I see from textbook that prime number must be irreducible number and not vice versa. But my degree of awareness is limitted in number field, so it's hard to understand because irreducible number is equivalent to prime number in number field. Thanks for helping me!
$\endgroup$ 21 Answer
$\begingroup$More accurately, the concept of "prime" is not equivalent to the concept of "irreducible". Irrreduciblity is a generalization of primality, every prime is irreducible, but it is possible in some algebraic structures that an irreducible element is not prime. The trouble is, in some algebraic structures, they ARE equivalent (most importantly, unique factorization domains).
The definition of irreducible:
If $a = bc \implies b$ or $c$ is a unit.
The definition of prime:
$p|bc \implies p|b$ or $p|c$.
A counter-example to: irreducible $\implies$ prime.
Consider $\Bbb Z[\sqrt{-5}]$.
We first claim $3$ is irreducible. To do this, we introduce the norm function:
$N(a + b\sqrt{-5}) = a^2 + 5b^2$.
It can be shown (I urge you to do this), that:
if $u = a+b\sqrt{-5}, v = c + d\sqrt{-5}$, that $N(uv) = N(u)N(v)$.
Also, if $u \neq \pm 1$, then $N(u) \geq 4$ (why?).
If $u$ is to be a unit, so it has an inverse $u^{-1}$, we have:
$1 = N(1) = N(uu^{-1}) = N(u)N(u^{-1})$. So a unit has to have a norm of $1$, since the norm is a positive integer. Thus the only units are $\pm 1$.
Now $N(3) = 9$, so if we have a factorization of $3$ into non-units, say:
$3 = xy$, then we must have $N(ax) = N(by) = 3$. But no element of $\Bbb Z[\sqrt{-5}]$ has a norm of $3$ (see above).
Now $9 = (2 + \sqrt{-5})(2 - \sqrt{-5})$, and $3|9$, but $3\not\mid(2 + \sqrt{-5})$ and $3\not\mid(2-\sqrt{-5})$, so $3$ is not prime.
$\endgroup$
